It is proven that for any a set of $n+1$ points, there exists a unique polynomial of degree at most $n$ that satisfies the points.
When is it the case that for a set of $n+1$ points, the unique polynomial that satisfies the points has exactly degree n?
On
The polynomial will have degree exactly $n$ if and only if the following condition is satisfied:
$$ 0 \ne \sum_{k=0}^n\frac{y_k}{\prod_{j\ne k}(x_k-x_j)}$$
To elaborate, suppose $(x_0,y_0),\ldots,(x_n,y_n)$ are the points through which we would like our polynomial to pass. The general model for a polynomial is $$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$$
We can write the system of equations $f(x_i) = y_i$ as $Va=y$, where $V$ is the Vandermonde matrix from $x_0,\ldots, x_n$, $a=(a_0,\ldots ,a_n)^\intercal$, and $y=(y_0,\ldots ,y_n)^\intercal$. Explicitly, $$ \left(\begin{array}{ccccc} x_0^n & x_0^{n-1} & \ldots & x_0 & 1 \\ x_1^n & x_1^{n-1} & \ldots & x_1 & 1 \\ \vdots & &\ddots&& \\ x_n^n & x_n^{n-1} & \ldots & x_n & 1 \\ \end{array}\right)\left(\begin{array}{c}a_n \\ a_{n-1} \\ \vdots \\ a_1 \\ a_0\end{array}\right) = \left(\begin{array}{c}y_0 \\ y_1 \\ \vdots \\ y_{n-1}\\ y_n\end{array}\right) $$ If we want a degree $n$ polynomial, we need $a_n\ne 0$. To get $a_n$, we compute $V^{-1}y$ and take the first entry.
We can come up with the polynomial: $f(x)=\displaystyle\sum_{i=0}^nf(a_i)\displaystyle\prod_{j\neq i}\frac{x-a_j}{a_i-a_j}$.
The leading coefficient must not be $0$ in order for it to have degree $n$ exactly; but the leading coefficient is $\displaystyle\sum_{i=0}^n\frac{f(a_i)}{\displaystyle\prod_{j\neq i}(a_i-a_j)}$. So we require that this is nonzero.
For example, if $a_i=i$ for i between $0$ and $n$, then $\displaystyle\prod_{j\neq i}(a_i-a_j)=i!(n-i)!(-1)^i$, so multiplying by $n!$, we see that such a polynomial is degree $n$ If and only if $f(0)-\binom{n}{1}f(1)+\binom{n}{2}f(2)-\ldots+(-1)^nf(n)\neq0$.